At heart, you need this version of the division algorithm to even define $(a\bmod m).$
Let $a,m\in\mathbb Z, m\neq 0.$ Then there is a unique pair $q,r\in \mathbb Z$ such that $a=mq+r$ and $0\leq r<|m|.$
Often, division algorithm does not include uniqueness. Depending on whether you’ve already proven division algorithm this way, you might have to prove a corollary to get uniqueness.
From that theorem, you define $(a\bmod m):=r,$ since $r$ is unique. Without uniqueness, you can’t even define $a\bmod m.$
But then $a-(a\bmod m)=a-r=mq$ is divisible by $m,$ so $$a\equiv (a\bmod m)\pmod m,$$ by definition.
That assumes you’ve defined congruence the easiest and most usual way:
$x\equiv y\pmod m$ iff $x-y$ is divisible by $m.$
Your approach would seem to indicate a slightly harder definition:
$x\equiv y\pmod m$ iff $(x\bmod m)=(y\bmod m).$
Then you would want to show: $$((a\bmod m)\bmod m)=(a\bmod m).\tag1$$
That follows from:
Lemma: If $0\leq c<|m|,$ then $(c\bmod m)=c.$
Proof: Eince $c=m\cdot 0+c$ satisfies the division algorithm condition, with $q=0, r=c,$ you are done.
Then, by definition of $a\bmod m,$ you’d have $$0\leq (a\bmod m)<|m|,$$ and hence you can conclude $(1)$ from the Lemma.
As you can see, all the work is really in the definitions, and which definitions you choose.
In first order logic formalism, we can’t even define new terminology. We have to replace all terms like “$a\bmod m$“ and “$x\equiv y\pmod m$” and “$|m|$” and “$k$ is divisible by $m$“ by their definitions.
In Peano Axioms for the natural numbers (non-negative integers,) we can’t even talk about “$x-y$” in general.
Nobody wants to do that.